Higher Engineering Mathematics

(Greg DeLong) #1
MACLAURIN’S SERIES 71

A


  1. Use Maclaurin’s series to determine the
    expansion of (3+ 2 t)^4.
    [
    81 + 216 t+ 216 t^2 + 96 t^3 + 16 t^4


]

8.5 Numerical integration using


Maclaurin’s series


The value of many integrals cannot be determined
using the various analytical methods. In Chapter
45, the trapezoidal, mid-ordinate and Simpson’s
rules are used to numerically evaluate such inte-
grals. Another method of finding the approximate
value of a definite integral is to express the func-
tion as a power series using Maclaurin’s series, and
then integrating each algebraic term in turn. This is
demonstrated in the following worked problems.


Problem 11. Evaluate

∫ 0. 4
0. 1 2e

sinθdθ, correct
to 3 significant figures.

A power series for esinθ is firstly obtained using
Maclaurin’s series.


f(θ)=esinθ f(0)=esin 0=e^0 = 1

f′(θ)=cosθesinθ f′(0)=cos 0 esin 0=(1)e^0 = 1

f′′(θ)=(cosθ)(cosθesinθ)+(esinθ)(−sinθ),
by the product rule,

=esinθ(cos^2 θ−sinθ);

f′′(0)=e^0 (cos^20 −sin 0)= 1

f′′′(θ)=(esinθ)[(2 cosθ(−sinθ)−cosθ)]

+(cos^2 θ−sinθ)(cosθesinθ)

=esinθcosθ[−2 sinθ− 1 +cos^2 θ−sinθ]

f′′′(0)=e^0 cos 0[(0− 1 + 1 −0)]= 0


Hence from equation (5):


esinθ=f(0)+θf′(0)+


θ^2
2!

f′′(0)+

θ^3
3!

f′′′(0)+···

= 1 +θ+

θ^2
2

+ 0

Thus

∫ 0. 4

0. 1

2esinθdθ=

∫ 0. 4

0. 1

2

(
1 +θ+

θ^2
2

)

=

∫ 0. 4

0. 1

(2+ 2 θ+θ^2 )dθ

=

[
2 θ+

2 θ^2
2

+

θ^3
3

] 0. 4

0. 1

=

(
0. 8 +(0.4)^2 +

(0.4)^3
3

)


(
0. 2 +(0.1)^2 +

(0.1)^3
3

)

= 0. 98133 − 0. 21033
= 0. 771 , correct to 3 significant figures

Problem 12. Evaluate

∫ 1

0

sinθ
θ

dθ using

Maclaurin’s series, correct to 3 significant
figures.

Let f(θ)=sinθ f(0)= 0
f′(θ)=cosθ f′(0)= 1
f′′(θ)=−sinθ f′′(0)= 0
f′′′(θ)=−cosθ f′′′(0)=− 1

fiv(θ)=sinθ fiv(0)= 0
fv(θ)=cosθ fv(0)= 1

Hence from equation (5):

sinθ=f(0)+θf′(0)+

θ^2
2!

f′′(0)+

θ^3
3!

f′′′(0)

+

θ^4
4!

fiv(0)+

θ^5
5!

fv(0)+···

= 0 +θ(1)+

θ^2
2!

(0)+

θ^3
3!

(−1)

+

θ^4
4!

(0)+

θ^5
5!

(1)+···

i.e. sinθ=θ−

θ^3
3!

+

θ^5
5!

−···

Hence

∫ 1

0

sinθ
θ

dθ=

∫ 1

0

(

θ−

θ^3
3!

+

θ^5
5!


θ^7
7!

+···

)

θ


=

∫ 1

0

(
1 −

θ^2
6

+

θ^4
120


θ^6
5040

+···

)
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